# Wood structures: story shear & unit shear question

1. Nov 27, 2007

### mrmotobiker

1. The problem statement, all variables and given/known data

A rectangular two-story building with a wood roof and floor system and masonry walls is analyzed to determine seismic forces.

Given: I=1, roof dead load D=15psf, floor load=25psf including partition load. Exterior wall dead load D=80 psf along both long and short walls. Ta=0.189seconds. SDS=0.4g and SD1=0.145. Redundancy factor is 1.0. R=5 for a bearing wall system with special reinforced masonry shearwalls.

The story height is 12 ft each floor. Additional 2ft of parapet is above the roof. Assume the masonry walls are along both long and short direction. The story dimension is 100'x40'

FIND:
a.) Determine the story shear on shear wall between roof and second-floor level along short walls.
b.) Determine the unit shear in roof diaphragm along short walls.

2. Relevant equations

Cs=SDSI/R=0.08g (given)
k=1 because 0.189s < 0.5s

Fx Coefficient=(0.08*608)wi*hi=0.00478

3. The attempt at a solution

a.)

Weight of Roof:
1ft*40'*15psf+(2'+12'/2)*80psf*2=1880plf
Tributary Weight of Roof: 1880plf*100'+(12'/2+2')(80psf)(40')*2=239.2kips

Weight of 2nd Floor:
1ft*25psf*40'+2*(12'/2+12'/2)*80psf=2920plf
Tributary Weight of 2nd floor: 2920plf*100'+2(12'/2+12'/2)*40'*80psf=368.8kips

Fx(ROOF)=Fxcoeff*height=0.00478*24'=0.115
Fx(2nd)=Fxcoeff*height=0.00478*12;=0.057

Roof Reaction:
1/2*0.115*1880plf*100'=10.81kips
2nd Floor Reaction:
1/2*0.057*2920pcf*100=8.322kips

Mid-story shear=10.81+(0.115*80*(12;/2+2)*40)=13.754kips

b.)

Weight of Roof=2920plf
Wupr=(0.115)*2920=335.8 lb/ft
Vur=(Wupr*L)/2=(335.8*100)/2=16790 lbs
vur=Vur/b=16790/40=419.75 lb/ft

Could anyone please check if I'm doing this correctly? I'm very uneasy about this answer, not because it doesn't sound right, but I'd really like to have this correct or find out what I'm doing wrong. Thanks ahead of time!

2. Nov 29, 2007

### mrmotobiker

bump. this is done using ASD in california.

3. Dec 1, 2007

### Pyrrhus

Hey mrmotobiker,

My wood structure design is a little rusty, but i'll give it a try.

So far your calculations look correct. I might that for the tributary second floor loas a partition load is usually added about 10 psf for seismic design force, but maybe you don't have to add it.

So far your a) part looks ok. Another method for obtaining the shear at midheight of second-story walls only for a simple rectangular building with two equal transverse shearwalls is by dividing the total roof shear by 2.

Now part b) looks weird to me.

4. Dec 14, 2007

### mrmotobiker

thanks. i got part a right, used the wrong weight for part b.

5. Dec 17, 2007

### mrmotobiker

hey cyclovenom, i realized we're the same age (your profile) lol. what school do you go to? i'm really looking into grad school for structural as well.

6. Dec 17, 2007

### Pyrrhus

Hey motorbiker, i'm dominican and i studied (graduated already) at http://www.unibe.edu.do" [Broken] . I'm planning on going to GATech on 2008 summer.

Last edited by a moderator: May 3, 2017